Simplifying the expression $\sqrt{(1+\cos t)^2+(-\sin t)^2}$

Question

I wish to find the length of the parametric curve $$x = f(t) = t + \sin t, y = g(t) = \cos t, t \in [0, \pi]$$

The length $L$ is given by $$\int_0^\pi\sqrt{(f'(t))^2+(g'(t))^2}dt = \int_0^\pi\sqrt{(1+\cos t)^2 + (-\sin t)^2}dt$$

Now, what I am able to do with this expression is to simplify it to $$\sqrt{1+2\cos t + \cos^2 t + \sin^2 t}$$ The identity $\sin^2 t + \cos^2 t = 1$ is staring at me, but I do not see how $$\sqrt{2\cos t + 2} = \sqrt{2}\sqrt{\cos t + 1}$$ is going to help me evaluate the integral. Is there another way to simplify the expression so that I'll get an easily computable integral? This task is early in Calc 2, so no very tricky tricks should be needed. I assume I need some expression with [trigonometric identity]-squared so that I will get rid of the square-root, but I honestly don't see how.

Answer 1

By the double angle formula,

$$\cos{t} = 2\cos^2\frac{t}{2} - 1$$

So your last line simplifies into $$\sqrt{2\cdot2\cos^2{\frac{t}{2}}}\\=2\left|\cos\frac{t}{2}\right|$$

Answer 2

$$ (1+\cos t)^2+(-\sin t)^2=1+2\cos t+\cos^2 t+\sin^2 t=2+2\cos t=4\cos^2\big(\tfrac{t}{2}\big). $$ Hence $$ \sqrt{(1+\cos t)^2+(-\sin t)^2}=2\big\lvert\cos\big(\tfrac{t}{2}\big)\big\rvert. $$